Add business days

It starts out as a fairly simple, innocent business requirement. Create a report to list records meeting some criteria, one of which is:

“List only records where today’s date is more than 35 business days after the due date from the record.”

When you delve deeper you find that querying the table with “DUE_DATE + 35 < SYSDATE” is not going to cut it – “business days” do not include weekends. You might start with something similar to this. But even that’s not good enough, because business days should not include public holidays. How do you code that?

So, here’s my solution.

1. We need to know what days are public holidays for the region. In our case this application is only applicable for a single region, so we use a simple table:

CREATE TABLE holidays (holiday_date DATE PRIMARY KEY);

We create a simple form for users to enter new holidays every year, and give someone the job of making sure it’s up-to-date every year when the public holidays are announced.

2. Create a view that lists all non-business days – i.e. list all weekends and public holidays. To retain reasonable performance, we limit our solution to dates in the years 2000 to 2050.

3. Now, when we need to take a date and add x business days to it, we query this table to find all the non-business-days that are applicable, e.g.:

SELECT day
,COUNT(*) OVER (ORDER BY day
ROWS BETWEEN UNBOUNDED PRECEDING
AND CURRENT ROW)
AS count_so_far
,(day - p_date) AS base_days
FROM NON_BUSINESS_DAYS
WHERE day > p_date;

If you run this query and examine each row in order of day, if you take base_days and subtract count_so_far, when the result is less than x, then base_days – count_so_far is the number of extra days we need to add to the holiday’s date to give us the answer. You’ll find this logic in the function below.

In our final solution, we’ll also need to UNION in the date parameter as well, for the case where there are no holidays between the starting date and the number of business days requested.

Here’s our function to take any date (at least, any date between 2000 and 2050) and add x business days (positive or negative):